Question

Three urns $A, B$ and $C$ contain $4$ red, $6$ black; $5$ red, $5$ black; and $\lambda$ red, $4$ black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn $C$ is $0.4$ then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^2 = \lambda x$ with one vertex at the vertex of the parabola, is

Answer 1

$ P \left(\frac{ C }{ R }\right)=\frac{ P ( C ) P \left(\frac{ R }{ C }\right)}{ P ( A ) P \left(\frac{ R }{ A }\right)+ P ( B ) P \left(\frac{ R }{ B }\right)+ P ( C ) P \left(\frac{ R }{ C }\right)} $
$0.4=\frac{\frac{1}{3} \times \frac{\lambda}{(\lambda+4)}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \frac{\lambda}{(\lambda+4)}} $
$ \Rightarrow \lambda=6$

$\tan 30^{\circ}=3 t=\frac{3}{2} t^2$
$ \frac{1}{\sqrt{3}}=\frac{2}{t} $
$ t=2 \sqrt{3}$
$ \left(\frac{3}{2} t ^2, 3 t \right)=(18,6 \sqrt{3}) $
$ \ell^2=18^2+(6 \sqrt{3})^2 $
$ =324+108$
$ =432$